Question 1000536
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Use the Rational Roots Test.  If a polynomial function has a rational zero, it will be of the form *[tex \Large \pm\frac{p}{q}] where *[tex \Large p] is a factor of the constant term and *[tex \Large q] is a factor of the lead coefficient.


Using the Rational Roots Test, create your list of possible zeros.  Counting both positive and negative possibilities you have 24 possible rational zeros for your polynomial. The other two zeros are complex.


Then start testing them using Synthetic Division.  I'll save you a little work: once you have found two real zeros, you have found them all.  The other two zeros are complex.


If you need a refresher on Synthetic Division, look here:


<a href="http://www.purplemath.com/modules/synthdiv.htm">Purple Math Synthetic Division</a>


Note that there are 4 pages; read them all.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \